Question

The motors${\mathbf{M}}_{\mathbf{1}}$and ${\mathbf{M}}_{\mathbf{2}}$of Problem 3.45 have inputs of 120 MWand 60 MW, respectively, at 13.2 kV, and both operate at unity power factor. Determine the generator terminal voltage and voltage regulation of the line. Neglect transformer phase shifts.

Short answer

The voltage regulation is 0.0255 and the voltage at generator terminal is $V_G=19.652\mathrm{kV}$.

Step-by-step solution

Given data

The power input to motors $M_1$and $M_2$ is 120 MW and 60 MW .

Determine the formulas ofper-unit voltage, power and current.

Write the formula of per-unit voltage.

${\mathbf{V}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{actual}}}{{\mathbf{V}}_{\mathbf{base}}}$……. (1)

Write the formula of per-unit power

${\mathbf{S}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{original}}}{{\mathbf{S}}_{\mathbf{base}}}$ ……. (2)

Write the formula of per-unit power

${\mathbf{I}}_{\mathbf{pu}}\frac{{\mathbf{S}}_{\mathbf{pu}}}{{\mathbf{V}}_{\mathbf{pu}}}$ ……. (3)

Determine the per-unit current through motor.

The base voltage at motor bus is $V_{base}=13.8\mathrm{kV}$and the actual voltage is $V_{actual}=13.2\mathrm{kV}$.

So, the per-unit voltage using equation (1) is calculated as,

$$\begin{gathered} V_{pu}=\frac{13.2\mathrm{kV}}{13.8\mathrm{kV}} \\ =0.9565\mathrm{pu} \end{gathered}$$

The base power in system is $S_{base}=300\mathrm{MW}$and the actual power is $S_{actual}=\left(120+60\right)\mathrm{MW}$.

So, the per-unit power using equation (2) is calculated as,

$$\begin{gathered} S_{pu}=\frac{\left(120+60\right)\mathrm{MW}}{300\mathrm{MW}} \\ =0.6\mathrm{pu} \end{gathered}$$

Substitute $0.6\mathrm{pu}\mathrm{for}{S}_{pu}\mathrm{and}0.9565\mathrm{pu}\mathrm{for}{V}_{pu}$in equation (3).

$$\begin{gathered} I_{pu}=\frac{0.6\mathrm{pu}}{0.9565\mathrm{pu}} \\ =0.6273\mathrm{pu} \end{gathered}$$

Determine the voltage regulation and terminal voltage at generator.

The per-unit impedance of transformer 2 from single-line diagram of problem 3.45 is $X_{T2}=0.0915\mathrm{pu}$, similarly, for transmission line is $X_{line}=0.0915\mathrm{pu}$and transformer 1 is $X_{T1}=0.0857\mathrm{pu}$.

Determine the voltage drop at point m for the above system.

$$\begin{gathered} V_1=V_{pu}+j{X}_{T2}{I}_{pu} \\ =0.9565+j\left(0.0915\right)\left(0.6273\right) \\ =0.9582\angle 3.434°\mathrm{pu} \end{gathered}$$

Determine the voltage drop at point k for the above system.

$$\begin{gathered} V_2=V_{pu}+j\left(X_{T2}+X_{T1}+X_{line}\right)I_{pu} \\ =0.9565+j\left(0.0915+0.1815+0.0857\right)\left(0.6273\right) \\ =0.9826\angle 13.237°\mathrm{pu} \end{gathered}$$

The voltage regulation is calculated as,